Question

求旋转抛物面z=x^2+y^2,平面z=0与圆柱面x^2+y^2=ax所围立方体的体积

Answer 1

作变换x=rcosu,y=rsinu,设D:x^2+y^2<=ax(a>0),
旋转抛物面z=x^2+y^2,平面z=0与圆柱面x^2+y^2=ax所围立方体的体积
=∫∫<D>(x^2+y^2)dxdy
=∫<-π/2，π/2>du∫<0,acosu>r^3dr
=∫<-π/2，π/2>(1/4)(acosu)^4du
=(a^4/8)∫<0,π/2>(1+cos2u)^2du
=(a^4/8)∫<0,π/2>(1+2cos2u+cos^2u)du
=(a^4/8)∫<0,π/2>[3/2+2cos2u+(1/2)cos4u]du
=3πa^4/32.